Expected value of quotient of Poisson distributions

Let $$X$$ and $$Y$$ be independent random variables such that $$X \sim \text{Poisson}(\lambda \cdot c)$$ and $$Y \sim \text{Poisson}(\lambda \cdot (1-c))$$, where $$c$$ is a real number in $$[0, 1]$$.

Is there an easy way of proving that $$E\left[\frac{X}{X + Y} | X + Y > 0\right] = c$$? Numerical computations I made indicate that this is true.

• Google "Poisson thinning." – whuber Nov 14 '18 at 15:13
• Aha! We can define Z := X + Y and see X as a thinning of Z with probability c. Then given any outcome of Z, conditional on Z, X is binomially distributed with parameters (Z, c), which has expected value c* Z. So for each Z!=0, E[X/Z | Z] = c. So E[X/Z | Z > 0] = c. Correct? – Christian Nov 14 '18 at 16:47